Question

For a certain metal, when monochromatic light of wavelength \(\lambda\) is incident, the stopping potential for photoelectrons is \(3V_0\). When the same metal is illuminated by light of wavelength \(2\lambda\), the stopping potential becomes \(V_0\). The threshold wavelength for photoelectric emission for the given metal is \(\alpha \lambda\). The value of \(\alpha\) is:

• A. \(1\)
• B. \(4\)
• C. \(2\)
• D. \(3\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Einstein's photoelectric equation relates the energy of incident photons to the work function of the metal and the maximum kinetic energy (stopping potential) of the emitted electrons.
Step 2: Key Formula or Approach:
Photoelectric equation: \[ eV_s = \frac{hc}{\lambda} - \phi = \frac{hc}{\lambda} - \frac{hc}{\lambda_{th}} \]
Step 3: Detailed Explanation:
For incident light of wavelength \(\lambda\):
\[ 3eV_o = \frac{hc}{\lambda} - \phi \quad \dots \text{(Eq. 1)} \]
For incident light of wavelength \(2\lambda\):
\[ eV_o = \frac{hc}{2\lambda} - \phi \quad \dots \text{(Eq. 2)} \]
Multiply (Eq. 2) by 3 to eliminate \(V_o\):
\[ 3eV_o = \frac{3hc}{2\lambda} - 3\phi \quad \dots \text{(Eq. 3)} \]
Equating (Eq. 1) and (Eq. 3):
\[ \frac{hc}{\lambda} - \phi = \frac{3hc}{2\lambda} - 3\phi \]
\[ 3\phi - \phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} \]
\[ 2\phi = \frac{hc}{2\lambda} \implies \phi = \frac{hc}{4\lambda} \].
We know that the work function \(\phi = \frac{hc}{\lambda_{th}}\):
\[ \frac{hc}{\lambda_{th}} = \frac{hc}{4\lambda} \implies \lambda_{th} = 4\lambda \].
Comparing with \(\lambda_{th} = \alpha \lambda\), we get \(\alpha = 4\).
Step 4: Final Answer:
The value of \(\alpha\) is 4.